- HERE WE'RE GIVEN THE POLYNOMIAL FUNCTION D OF T =
8T TO THE FOURTH - 36T TO THE THIRD - 22 SQUARED.
WE'RE ASKED, WHAT IS THE D INTERCEPT
AND WHAT ARE THE T INTERCEPTS.
WHEN USING FUNCTIONS,
THE VARIABLES USED ARE OFTEN NOT X AND Y LIKE WE'RE USED TO,
BUT THE PROCESS FOR FINDING THE INTERCEPTS IS THE SAME.
IN GENERAL, IF WE HAVE A FUNCTION
TO DETERMINE THE HORIZONTAL INTERCEPTS,
OFTEN REFERRED TO AS THE X INTERCEPTS,
WE SET F OF X EQUAL TO 0 WHERE THE FUNCTION EQUAL TO 0
AND SOLVE FOR X.
AND TO DETERMINE THE VERTICAL INTERCEPT OR THE Y INTERCEPT,
WE SET X = TO 0 OR EVALUATE F OF 0.
SO WHEN WE'RE ASKED TO FIND THE D INTERCEPT,
REMEMBER D OF T WOULD BE = TO Y,
SO THIS IS THE SAME AS THE Y INTERCEPT OR VERTICAL INTERCEPT.
WE'RE GOING TO SET THE INPUT, IN THIS CASE T = 0,
AND EVALUATE THE FUNCTION.
SO TO FIND THE D INTERCEPT, WE SET T = 0, OR THE INPUT = 0.
SO WE'D HAVE D OF 0 = 8 x 0 TO THE FOURTH
MINUS 36 0 TO THE THIRD MINUS 20 0 SQUARED,
SO THIS IS EQUAL TO 0.
SO IF D OF 0 = 0 THIS TELLS US OUR D INTERCEPT IS THE ORIGIN,
SO THE COORDINATES ARE (0,0).
TO FIND THE T INTERCEPTS,
THIS WOULD BE THE SAME AS FINDING THE X INTERCEPTS.
WE'RE GOING TO SET Y,
OR IN THIS CASE THE FUNCTION EQUAL TO ZERO
AND SOLVE FOR T.
SO WE'RE GOING TO SET THE FUNCTION D OF T = 0
AND SOLVE FOR T.
SO WE'RE GOING TO SOLVE THE EQUATION 0 = 8T TO THE FOURTH
MINUS 36T TO THE THIRD MINUS 20T SQUARED.
LUCKILY THIS IS FACTORABLE.
WE START BY FACTORING OUT THE GREATEST COMMON FACTOR,
WHICH IN THIS CASE WOULD BE 4T TO THE SECOND,
LEAVING US WITH 2T SQUARED - 9T - 5.
EVEN THOUGH OUR LEADING COEFFICIENT HERE IS 2,
WE CAN STILL FACTOR THIS.
FACTORS OF 2T SQUARED ARE 2T AND T.
WE WANT THE FACTORS OF -5
SO THE SUM OF THE INNER AND OUTER PRODUCT IS = TO -9T.
SO PUT A FACTOR OF -5 HERE AND -5, +1 HERE.
YOU HAVE -10T + 1T THAT IS EQUAL -9T,
SO THIS IS FACTORED CORRECTLY.
NOW, YOU SET EACH FACTOR EQUAL TO ZERO AND SOLVE FOR T.
SO WE CAN SEE HERE T IS EQUAL TO 0 FROM THESE TWO FACTORS.
IF WE SET 2T + 1 = 0, SUBTRACT 1 ON BOTH SIDES, DIVIDE BY 2,
SO WE HAVE T = -1/2.
AND THEN T + 5 = 0 WHEN T = 5.
SO WE HAVE THREE T INTERCEPTS.
WE HAVE THE POINTS (0,0), (- 1/2,0), AND (5,0).
REMEMBER FOR THE T INTERCEPTS THE Y COORDINATE,
OR IN THIS CASE THE D COORDINATE,
IS EQUAL TO 0 EACH TIME.
LET'S GO AHEAD AND VERIFY OUR WORK
BY ANALYZING THE GRAPH OF OUR FUNCTION.
THIS ISN'T A GREAT WINDOW FOR OUR GRAPH
BECAUSE REMEMBER WE SET THE HORIZONTAL INTERCEPTS,
OR THE T INTERCEPTS, WHERE -1/2, 0, AND +5.
WE CAN'T REALLY TELL RIGHT HERE THERE ARE TWO T INTERCEPTS.
AND WE ALSO SAID THE D INTERCEPT WAS THE ORIGIN,
WHICH WE'VE ALREADY HIGHLIGHTED HERE.
I HOPE YOU FOUND THIS HELPFUL.